A charge having $q/m$ equal to $10^8\, C/kg$ and with velocity $3 \times 10^5\, m/s$ enters into a uniform magnetic field $0.3\, tesla$ at an angle $30^o$ with direction of field. The radius of curvature will be ......$cm$
AIPMT 2000, Medium
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$r = \frac{{mv\sin \theta }}{{qB}}$
$r =\left|\frac{ m }{ q }\right|\left|\frac{3 \times 10^{5} \times \sin 30^{\circ}}{0.3}\right|$
$r =\frac{3 \times 105}{10^{8} \times 0.3 \times 2}=0.5 \times 10^{-2} m =0.5 cm$
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